# Does the almost sure convergence of absolutely continuous r.v.'s imply the weak convergence of the pdf's in $(L^\infty)^*$?

The following question was asked in a comment at Almost sure convergence vs convergence of probability density functions :

Suppose that $$(X_n)$$ is a sequence of random variables (r.v.'s) converging almost surely (a.s.) to a r.v. $$X$$. Suppose that $$X_n$$ and $$X$$ have pdf's $$p_n$$ and $$p$$, respectively. Does it then necessarily follow that $$\int fp_n\to\int fp$$ for all $$f\in L^\infty[0,1]$$ (and $$n\to\infty$$)?

Here this question will be answered.

• Remark: Convergence does hold if f is Riemann integrable, i.e., continuous a.e. Feb 9, 2021 at 16:28

The answer is no. Indeed, let $$C$$ be any countable subset of $$(0,1)$$ which is dense in $$[0,1]$$. Write $$C=\{c_1,c_2,\dots\}$$, where the $$c_n$$'s are pairwise distinct. Let $$U:=(0,1)\cap\bigcup_{n=1}^\infty(c_n+4^{-n}B)\quad\text{and}\quad F:=[0,1]\setminus U,$$ where $$B:=(-1,1)$$. Then $$U\supset S$$ and hence $$U$$ is dense in $$[0,1]$$. The Lebesgue measure $$|U|$$ of $$U$$ is $$\le2/3$$ and hence $$|F|\ge1/3>0$$. Let now $$X$$ be any r.v. uniformly distributed on $$F$$, so that the pdf of $$X$$ is $$p=1_F/|F|.$$

Also, the set $$U$$ is open and therefore $$U=\bigcup_{k=1}^\infty I_k,$$ where the $$I_k$$'s are some pairwise disjoint open subintervals of $$(0,1)$$. Let $$U_n:=\bigcup_{k=1}^n I_k$$ and then let $$E_n$$ denote the set of all endpoints of the intervals $$I_1,\dots,I_n$$. For any $$x\in[0,1]$$ and any $$A\subseteq[0,1]$$, let $$d(x,A):=\inf\{|y-x|\colon y\in A\}$$.

Since $$U$$ is dense in $$[0,1]$$, for all $$x\in F$$ we have $$d(x,U)=0$$ and hence $$d(x,U_n)\to0$$, so that $$d(x,E_n)=d(x,U_n)\to0$$. Letting now $$Y_n:=y_n(X),$$ where $$y_n(x):=\min\{y\in E_n\colon |x-y|=d(x,E_n)\},$$ we see that $$|X-Y_n|=d(X,E_n)\to 0$$ a.s. and hence $$Y_n\to X$$ a.s.

Finally, define $$X_n$$ as follows. For any $$y\in E_n$$ and any $$t\in(0,1)$$, let

(i) $$x_n(y,t):=y+|I_k|t/n$$ if $$y$$ is the left endpoint of the interval $$I_k$$ for some $$k\in[n]:=\{1,\dots,n\}$$, and

(ii) $$x_n(y,t):=y-|I_k|t/n$$ if $$y$$ is the right endpoint of the interval $$I_k$$ for some $$k\in[n]$$ but $$y$$ is not the left endpoint of the interval $$I_l$$ for any $$l\in[n]$$.

Then let $$X_n:=x_n(Y_n,T)$$, where $$T$$ is a r.v. uniformly distributed on $$(0,1)$$ and independent of $$X$$ (and hence of $$Y_n$$). Then $$X_n$$ is an absolutely continuous r.v. with all values in $$U_n\subseteq U$$; as before, let $$p_n$$ denote the pdf of $$X_n$$. Also, $$|X_n-Y_n|\le1/n$$. Therefore and because $$Y_n\to X$$ a.s., we have $$X_n\to X$$ a.s. However, $$1_U\in L^\infty[0,1]$$ but $$\int 1_U\,p_n=P(X_n\in U)=1\not\to0=P(X\in U)=\int 1_U\,p.$$

• Once $U$ and $X$ are defined, you can just define the conditional distribution of $X_n$ given $X$ to be uniformly distributed on the open set $(X-1/n,X+1/n) \cap U$. That will shorten the proof. Feb 9, 2021 at 3:50
• @YuvalPeres : Thank you for your comment. Feb 9, 2021 at 16:20
• Thank you very much, @IosifPinelis! I guess this shows that the convergence you gave is the strongest possible. Feb 14, 2021 at 12:35
• @NateRiver : I am glad you found this of help. So, to have a closure, are you satisfied with this answer? Feb 14, 2021 at 16:21
• Yes, is there anything I should do to mark this? I think I have accepted your answer in my original post. Feb 19, 2021 at 23:46